I have read that what differentiates an irreversible process like adiabatic free expansion from its counterpart reversible process — isothermal expansion, which takes the system to the same final state — is the change in entropy of the surroundings. In the free expansion case it is $0,$ whereas in isothermal expansion $\displaystyle \Delta S_\mathrm{surr} = -nR\ln\frac{V_2}{V_1}$ such that $\Delta S_\mathrm{tot} = 0$.

I am unable to understand why the entropy change of the surroundings in the free expansion case is $0$.

$$\Delta S_\mathrm{surr} = \int \frac{\mathrm dq_\mathrm{irrev}}{T_\mathrm b} + S_\mathrm{gen} = \int \frac{\mathrm dq_\mathrm{rev}}{T}$$

Since the entropy transfer term $\displaystyle \int \frac{\mathrm dq_\mathrm{irrev}}{T_\mathrm b}$ is $0$ as the process is adiabatic, for the net entropy to equate to $0$ the entropy generated must also be $0.$ But doesn't this imply that the surrounding is undergoing a reversible process?

Also, I am unable to think of a suitable reversible process for the surroundings which would take it to the same final state. Moreover, what does the change in entropy even signify for a vacuum?

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    $\begingroup$ That final equation you wrote for the entropy change of the surroundings is incorrect. This equation is for the entropy change of the system. $\endgroup$ Commented Apr 16 at 11:38
  • $\begingroup$ @ChetMiller Which part? Or if the entirety of it is incorrect could you please guide me on what would be the correct equation to use in this case. Could you also tell me how what ive written differs from how the entropy is written for the reservoir in page 222 of Fundamentals of Engineering Thermodynamics by Moran $\endgroup$ Commented Apr 16 at 14:03
  • $\begingroup$ The lhs should be entropy of the system, not the surroundings. Page 222 of the book does not have surroundings entropy change in it. The entropy change of the surroundings corresponding to this system entropy change is -Q/Tb, where Q is the actual heat transferred to the system (not reversible path heat). $\endgroup$ Commented Apr 16 at 18:45
  • $\begingroup$ Oh ok i think I understood now. Do we not consider entropy generation for surroundings in all cases due to the fact that like a reservoir, it has uniform temperature throughout and is therefore free from irreversabilities? Also isn't the entropy change of reservoir mentioned in page 222 (page 237 of the pdf) the change for surroundings? Thank you for taking the time to clarify my doubts $\endgroup$ Commented Apr 17 at 3:17
  • $\begingroup$ Yes, the surroundings are considered to have no irreversibilities, and consist of ideal constant temperature reservoirs. $\endgroup$ Commented Apr 17 at 10:32

1 Answer 1


Since entropy is a state variable, $\Delta S_\mathrm{sys}$ between two states (defined by variables such as T, p, V, n) is the same whether the process is carried out reversibly or irreversibly. According to the second law of thermodynamics, for the system

$$\Delta S_\mathrm{sys} = \int \frac{dq_{rev}}{T} $$

Again, the value $\Delta S_\mathrm{sys}$ will be the same for an irreversible process even though the heat exchanged during the process will be $\int dq$, not $\int dq_{rev}$.

For the surroundings $$\Delta S_\mathrm{surr} = -\int \frac{dq}{T}$$

If the surroundings consists of vacuum, its temperature is nominally absolute zero and entropy is zero (here according to the third law of thermodynamics).

  • $\begingroup$ So the reason of 0 change in surrounding entropy isnt due to the adiabatic nature of the process but because vacuum itself doesnt have an entropy associated to it ? $\endgroup$ Commented Apr 16 at 11:06
  • $\begingroup$ @JeffJefferson Both, I'd say. There isn't anything to change in the surroundings, nothing to receive heat (or energy, period). $\endgroup$
    – Buck Thorn
    Commented Apr 16 at 15:17

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